The Lebesgue measure $\lambda$ is a function on a subset of the power set of real numbers $\mathbb{R}$ that satisfies the following properties (among others):

(i) $\lambda$ is finitely additive: If $A$ and $B$ are disjoint measurable sets then $\lambda(A \cup B) = \lambda(A) + \lambda(B)$;
(ii) $\lambda$ is defined on a sigma algebra and is countably sub-additive: $\lambda\!\left(\bigcup_{i = 1}^{\infty} A_i\right) \leq \sum_{i = 1}^{\infty} \lambda(A_i)$;
(iii) $\lambda$ is translation invariant: $\lambda(A + c) = \lambda(A)$ for any constant $c$ where $A + c = \{a+c \mid a \in A\}$;
(iv) $\lambda$ respects scaling: $\lambda(cA) = c\lambda(A)$ for any constant $c$ where $cA = \{ca \mid a \in A\}$.

Whether there exist non-measurable sets depends on the assumed model of set theory. In the Solovay model--a model of ZF excluding C(hoice)--every subset of $\mathbb{R}$ is measurable. However, in ZF+C there exist sets that are not Lebesgue measurable (cf. Vitali set). I assume ZF+C for now.

Let us say that $g$ extends $f$ over $\mathbb{R}$ provided $g(A) = f(A)$ for all $f$-measurable sets $A$ and $g(B)$ is defined for at least one non-$f$-measurable set $B \subseteq \mathbb{R}$. I am interested in finding maximal functions (with respect to extension) that mostly satisfy the nice properties of the Lebesgue measure. Toward that end ...

Question: Which of the above properties can be dropped so that there exists some function in $ZF+C$ satisfying the other properties and extending the Lebesgue measure $\lambda$ over $\mathbb{R}$?

  • $\begingroup$ What is the definition of "measurable" in (i) and (v)? $\endgroup$
    – bof
    Feb 25, 2017 at 1:06
  • $\begingroup$ @bof In the domain of the function. $\endgroup$
    – D. Ror.
    Feb 26, 2017 at 4:03
  • $\begingroup$ That's what I thought. But doesn't (ii) say that the domain is a sigma algebra, and doesn't that imply (v)? $\endgroup$
    – bof
    Feb 26, 2017 at 4:16
  • $\begingroup$ @bof Good point. I'll remove (v) from the list to simplify. $\endgroup$
    – D. Ror.
    Feb 28, 2017 at 4:54

2 Answers 2


Let us deal with each weakening one-by-one:

Dropping (i): In this case Lebesgue outer measure is a total extension of Lebesgue measure that satisfies all other requirements.

Dropping (ii): Already addressed in Gerald's answer - There are scaling invariant (property (iv)) Banach measures on $\mathbb{R}$. (There are also non scaling invariant Banach measures.)

Dropping (iii): There is no such total extension $\mu$. To see this choose a basis $B$ of the vector space $(\mathbb{R}, +)$ over the field $\mathbb{Q}$ of rationals with $1 \in B$. Let $W$ be the $\mathbb{Q}$-linear span of $B \setminus \{1\}$ and $X = W + 1$. Then the rational multiples of $X$ are pairwise disjoint and cover $\mathbb{R}$. By (i) + (ii) = countable additivity and (iv), we must have $\mu(X) > 0$ and hence, $\mu(X \cap [n, n+1)) > 0$ for some integer $n$. Let $Y = X \cap [n, n+1)$ and $Y_k = (1/k)Y$ for $k \geq 1$. Then $\{Y_k: k \geq 1\}$ is a family of pairwise disjoint sets whose union is bounded and the sum of whose $\mu$-measures is infinite (by (iv)): Contradiction.

Dropping (iv): Vitali's example shows that no total extension exists.

Other results along these lines:

(1) (Solovay, 1971) The existence of a total extension of Lebesgue measure that satisfies (i) + (ii) (= countable additivity) is equiconsistent with the existence of a measurable cardinal.

(2) (Ciesielski-Pelc, 1985) There is no maximal isomoetric-invariant extension of Lebesgue measure.

  • 1
    $\begingroup$ Thank you! What about partial extensions when dropping (iii) or (iv)? $\endgroup$
    – D. Ror.
    Mar 1, 2017 at 16:09

If you drop (ii), then there is an extension to all subsets of $\mathbb R$.

The corresponding thing fails in $\mathbb R^3$, where (iii) is replaced by "rigid motions" --- translations, reflections, rotations.

  • 1
    $\begingroup$ Can you give any hints as to how to see this, or references? $\endgroup$ Feb 24, 2017 at 23:01
  • 1
    $\begingroup$ See Banach-Tarski paradox $\endgroup$ Feb 24, 2017 at 23:07
  • 2
    $\begingroup$ @NateEldredge: If you are asking for a reference for the first claim, see Exercise 2.15 in these notes. I think the result is originally due to Banach but I don't know the exact reference. $\endgroup$
    – Burak
    Feb 24, 2017 at 23:30
  • 1
    $\begingroup$ @Burak: Thanks! The semigroup-invariant Hahn-Banach theorem is new to me but very cool. $\endgroup$ Feb 24, 2017 at 23:36

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